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John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.35 s? Enter your answer to at least three significant digits. Assuming that all the numbers given are exact, what is John's position at a time of 4.35 s? Enter your answer to at least three significant digits.

John is rollerblading down a long straight path At time zero there is a mailbox about 1 m in front of him In the 5 s time period that follows Johns velocity is class=

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onyebuchinnaji

The position of John at time of 4.35 s is -1.13 m

From the given chart;

  • at time, t = 0, the initial velocity, u = - 2 m/s
  • at time, t = 5 s, the final velocity, v = 2 m/s

The average acceleration is calculated as follows;

[tex]acceleration = \frac{\Delta \ velocity}{\Delta \ time} \\\\acceleration = \frac{v- u}{t_2 - t_1} = \frac{2 - (-2)}{5 - 0} = \frac{4}{5} = 0.8 \ m/s^2[/tex]

The position of John at time, t = 4.35 s is calculated as follows;

[tex]x(t) = ut + \frac{1}{2} at^2\\\\x(4.35) = (-2 \times 4.35) \ + \ 0.5\times 0.8\times( 4.35)^2\\\\x(4.35) = -8.7 \ + \ \ 7.569\\\\x(4.35) = - 1.13 \ m[/tex]

Thus, the position of John at time of 4.35 s is -1.13 m

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